CN
2
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
L
1 tháng 5 2019
\(A\cdot2=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{256}\right)\cdot2\)
\(=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}\)
\(A\cdot2-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}...+\frac{1}{128}-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(A=1-\frac{1}{256}=\frac{255}{256}\)
1 tháng 5 2019
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^7}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(A=1-\frac{1}{2^8}\)
\(A=\frac{2^8-1}{2^8}\)
\(A=\frac{255}{256}\)
PN
1
15 tháng 11 2016
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
M
15 tháng 8 2017
ta có : A=1/2+1/4+..+1/1024
=> A=1/21+1/22+..+1/210
=> A.2=(1/21+1/22+..+1/210).2
=> A.2=1+1/21+1/22+..+1/29
=> 2A-A=(1+1/21+1/22+..+1/29)-(1/21+1/22+..+1/210)
=> A=1-1/210
5 tháng 2 2016
2y=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
2y-y=1-\(\frac{1}{256}\)
y=\(\frac{255}{256}\)
VH
2
N
15 tháng 7 2017
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}=\frac{481}{280}\)
27 tháng 8 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{64}{128}+\frac{32}{128}+\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}\)
\(=\frac{126}{128}=\frac{63}{64}\)
8 tháng 7 2017
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\)
\(=\frac{1+1+1+1+1+1+1}{2}\)
\(=\frac{7}{2}\)
DP
8 tháng 7 2017
Đặt \(T=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(T=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)\)
\(\Rightarrow T=1-\frac{1}{128}=\frac{127}{128}\)
8 tháng 7 2017
a, \(\frac{2}{3}+\frac{2}{3}+\frac{6}{3}=\frac{10}{3}\)
b,\(\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=\frac{6}{4}+\frac{3}{2}=\frac{3}{2}+\frac{3}{2}=\frac{6}{2}=3\)
Sửa lại là 1/256 nha
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{128}-\frac{1}{256}\right)\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)
\(=1-\frac{1}{256}=\frac{255}{256}\)
\(=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{128}-\frac{1}{256}\right)\)
\(=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{8}-\frac{1}{8}\right)-...-\left(\frac{1}{128}-\frac{1}{128}\right)-\frac{1}{256}\)
\(=1-\frac{1}{256}=\frac{255}{256}\)